How to make two gestures recognize at the same time using simultaneousGesture()

Paul Hudson @twostraws September 3rd 2021

Updated for Xcode 13.0

By default SwiftUI will trigger only one gesture recognizer action at a time, which is usually whichever one is the front-most view in your hierarchy – it would prefer a recognizer on a child view rather than its parent, for example. If you want to override this behavior to make two gestures trigger at once, you should use the simultaneousGesture() when creating your second and subsequent gestures.

For example, in this code we have two tap gestures, but SwiftUI will execute only the one attached to the circle because it’s the child of the VStack:

struct ContentView: View {
    var body: some View {
        VStack {
            Circle()
                .fill(Color.red)
                .frame(width: 200, height: 200)
                .onTapGesture {
                    print("Circle tapped")
                }
        }
        .onTapGesture {
            print("VStack tapped")
        }
    }
}

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If you want both gestures to trigger – i.e., if you want both “Circle tapped” and “VStack tapped” to be printed – you should use simultaneousGesture() on the VStack like this:

struct ContentView: View {
    var body: some View {
        VStack {
            Circle()
                .fill(Color.red)
                .frame(width: 200, height: 200)
                .onTapGesture {
                    print("Circle tapped")
                }
        }
        .simultaneousGesture(
            TapGesture()
                .onEnded { _ in
                    print("VStack tapped")
                }
        )
    }
}

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Note: You should use simultaneousGesture() with the gesture that would otherwise not be executed otherwise it won’t work. So, in our previous example using simultaneousGesture() with the circle and a simple onTapGesture() with the VStack will still print just “Circle tapped” – it won’t do what you expect.

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How to make two gestures recognize at the same time using simultaneousGesture()

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